In Tu's book, specifically the section on "Taylor's Theorem with remainder" page 7, Let be a ∞ function on an open set ⊆ℝ which is star shaped with respect to a point =(1,...,)∈. Then there are functions $_1()$,...,$_$()∈∞
such that
$()= f(p)+ ∑(x^i − p^i )g_i(x)$ , where $()=\frac{\\∂}{∂}(p)$.
by chain rule
$\frac{\\d}{dt}f(p+t(x−p))=∑(x^i−p^i)\frac{\\∂}{∂^i}(p+t(x−p))$.
If we integrate both sides with respect to t from 0 to 1, we get $ [f(p+t(x− p))]^1_0 = ∑(x^i − p^i) \int_0^1 \frac{\\∂}{∂^i} (p+t(x− p))dt$.
$g_i(x) = \int_0^1 \frac{\\∂}{∂^i}(p+t(x− p))dt$
Then gi(x) is C∞ and (1.1) becomes
$f(x)− f(p) = ∑(x^i − p^i)g_i(x)$.
$\frac{\\d}{dt}f(p+t(x−p))=∑(x^i−p^i)∂f(p+t(x−p))$.
Then later on he goes on to to say:
Moreover,
$g_i(p)=\int_0^1 \frac{\\∂}{∂}(p)dt$=$\frac{\\∂}{∂}(p)$
In case n=1and p=0,this lemma says that
$f(x) = f(0)+xg_1(x)$
for some C∞ function $g_1(x)$. Applying the lemma repeatedly gives
$g_i(x)=g_i(0)+xg_{i+1}(x)$
where $g_{i}$, $g_{i+1}$ are C∞ functions. Hence,
$f(x)= f(0)+x(g_{1}(0)+xg_{2}(x))$
= $f(0)+xg_{1}(0)+x^2(g_{2}(0)+xg_{3}(x))$
= $f(0)+xg_{1}(0)+x^2g_{2}(0)+...x^ig_{i}(0)+x^{i+1}g_{i+1}(x)$ (1.2)
I do not understand how he jumped from 1.2 to
Differentiating (1.2) repeatedly and evaluating at 0, we get
$g_k(0)=\ \frac{\\f^k(0)}{k!}$ $k=1,2,.....i.$
This is my first post so appolgies if i messed up the format.